To understand why the tension at 30° (\(T_1\)) is greater than at 60° (\(T_2\)), we need to examine the forces acting on a particle moving in a vertical circle.
When a particle moves in a vertical circle, the forces acting on it are gravitational force (\(mg\), where \(m\) is the particle's mass and \(g\) is the acceleration due to gravity) and the tension in the string. The tension provides the necessary centripetal force to keep the particle moving in the circle.
- At an angle \(\theta\) from the lowest point, the tension \(T\) in the string is given by: \(T = \frac{mv^2}{r} + mg \cos \theta\), where \(v\) is the velocity of the particle at that point, and \(r\) is the radius of the circle.
- Let's compare \(T_1\) at 30° and \(T_2\) at 60°:
- \(T_1 = \frac{mv_1^2}{r} + mg \cos 30^\circ = \frac{mv_1^2}{r} + \frac{\sqrt{3}}{2} mg\)
- \(T_2 = \frac{mv_2^2}{r} + mg \cos 60^\circ = \frac{mv_2^2}{r} + \frac{1}{2} mg\)
- Now, from energy conservation between these two points: \(\frac{1}{2} mv_1^2 + mg \cdot r(1 - \cos 30^\circ) = \frac{1}{2} mv_2^2 + mg \cdot r(1 - \cos 60^\circ)\)
- Simplifying the equation using trigonometric values:
- At 30°: Gain in potential energy = \(mgr\left(\frac{1}{2} - \frac{\sqrt{3}}{2}\right)\)
- At 60°: Gain in potential energy = \(mgr(1 - \frac{1}{2})\)
- Thus, the velocity at 30° (\(v_1\)) is greater than at 60° (\(v_2\)) due to greater potential energy drop at 60°, resulting in higher tension at 30° than at 60° because tension is directly related to the square of velocity.
Therefore, the correct statement is \(T_1 > T_2\) because tension decreases with an increase in angle due to the decrease in the component of the gravitational force (\(mg \cos \theta\)) and drop in velocity.
Hence, the correct answer is \(T_1 \gt T_2\).